40. Combination Sum II

Description

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target.

Each number in candidates may only be used once in the combination.

Note: The solution set must not contain duplicate combinations.

Example 1:

• Input: candidates = [10,1,2,7,6,1,5], target = 8
• Output:
  [
  [1,1,6],
  [1,2,5],
  [1,7],
  [2,6]
  ]

Example 2:

• Input: candidates = [2,5,2,1,2], target = 5
• Output:
  [
  [1,2,2],
  [5]
  ]

Constraints:

• 1 <= candidates.length <= 100
• 1 <= candidates[i] <= 50
• 1 <= target <= 30

Submitted Code

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        def dfs(start, total, comb):
            if total == target:
                res.append(comb.copy())
                return

            for i in range(start, len(candidates)):
                if i > start and candidates[i] == candidates[i-1]:  # 중복 숫자 스킵
                    continue
                if total + candidates[i] > target:                  # 그 뒤 조합 스킵
                    break

                comb.append(candidates[i])
                dfs(i + 1, total + candidates[i], comb)
                comb.pop()

        candidates.sort()
        res = []
        dfs(0, 0, [])

Runtime: 3 ms | Beats 86.66%
Memory: 19.54 MB | Beats 60.64%

39. Combination Sum 문제와 비슷하지만 이번에는 같은 숫자를 재사용할 수 없고 중복된 숫자가 존재한다.

Other Solutions

1st

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        candidates.sort()
        res = []

        def dfs(target, start, comb):
            if target < 0:
                return
            if target == 0:
                res.append(comb)
                return
            for i in range(start, len(candidates)):
                if i > start and candidates[i] == candidates[i-1]:
                    continue
                if candidates[i] > target:
                    break
                dfs(target-candidates[i], i+1, comb+[candidates[i]])

        dfs(target, 0, [])
        return res

time complexity: 𝑂(2^𝑛)
space complexity: 𝑂(𝑛)

이 방법은 매번 comb 리스트를 새로 생성하기 때문에 메모리는 더 사용하지만 append와 pop이 필요 없기 때문에 간단하다.